Derivative of $\arcsin(\cos(x))$ at $x=0$.

Question

We have to differentiate this function at $x=0$ $$\left. \dfrac{d \arcsin (\cos (x))}{dx} \right|_{x=0}$$ Using the identity $\cos x=\sin \Big(\dfrac{\pi}{2}-x\Big)$ we get $$\left. \dfrac{d \arcsin (\sin (\dfrac{\pi}{2} - x))}{dx} \right|_{x=0},$$ For $0 \le x \le \pi$. We know $\arcsin \Big(\sin \Big(\dfrac{\pi}{2}-x\Big)\Big)=\dfrac{\pi}{2}-x$.

So, our original problem reduces to $$\left. \dfrac{d\Big(\dfrac{\pi}{2}-x\Big)}{dx}\right|_{x=0},$$ which is equal to $-1$ for $0 \le x \le \pi$.

But the derivative of this function at $x=0$ is undefined. What's going on here?

Answer 1

Near zero you have $$\arcsin(\cos x) = \frac{\pi}{2} - \arccos(\cos x) = \frac{\pi}{2} - |x|$$ (so the derivative is $+1$ for $x<0$ and $-1$ for $x>0$). And therefore the function cannot be approximated by a linear function in a arbitrary small interval containg $0,$ and this implies that the derivative at $x=0$ does not exist.

Edit: Near $x = 0$ you have $\arccos(\cos x)=|x|$ because $\cos(x)$ is an even function. It cannot be just $x$ because the value is positive for $x\ne0$.

You also have to look at values $x<0,$ if you want to know whether the function has a derivative in a neighborhood of $0$ (it has, if the left and rights derivatives exist and are equal, see e.g. here).

Answer 2

The derivative of the function, for $x\ne0$, is $$ -\frac{\sin x}{\sqrt{1-\cos^2x}}=-\frac{\sin x}{\lvert\sin x\rvert} $$ By l'Hôpital, $$ \lim_{x\to0^+}\frac{\arcsin(\cos x)-\arcsin(\cos0)}{x}= \lim_{x\to0^+}-\frac{\sin x}{\lvert\sin x\rvert}=-1 $$ and, similarly, the limit from the left is $1$.

Thus the limit of the difference quotient doesn't exist.

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If we limit the function to the interval $(-\pi/2,\pi/2)$, we have that $$ f'(x)=\begin{cases} 1 & -\pi/2<x<0 \\[4px] -1 & 0<x<\pi/2 \end{cases} $$ so that $$ f(x)=\begin{cases} x+c_- & -\pi/2<x<0 \\[4px] -x+c_+ & 0<x<\pi/2 \end{cases} $$ for some constants $c_-$ and $c_+$; continuity at $0$ and $f(0)=\pi/2$ gives $c_-=c_+\pi/2$.

Now it should be clear what's your mistake: the equality $$ \arcsin(\sin(\pi/2-x)=\pi/2-x $$ doesn't hold for all $x$.